# Solving the equation $\frac{2x+3}{x+1} = \frac{2x+2}{x-1}$

I'm a college student and I've got a problem solving this:

$$\frac{2x+3}{x+1} = \frac{2x+2}{x-1}.$$

Since neither the numerators nor the denominators are equal, I figured that $(2x+3)(x-1) = (x+1)(2x+2)$ would provide me with a suitable $x$.

However calculating this I found that $x = -\frac13$ while my answer sheet states that $x = -\frac23$. What am I doing wrong?

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Are you certain you have written the problem correctly? My answer differs both from yours and from the answer sheet. – AMPerrine Oct 30 '11 at 20:09
Thank you for your answers everyone, I'm sorry about the answer from the answer sheet being wrong, it should have been -(5/3) – Henri Oct 30 '11 at 20:24
Hopefully this doesn't sound to condescending, but if you are ever unsure about a solution to an equation, often the easiest thing to do is just substitute the solution back in and see if everything works out. Indeed, it's probably not a bad practice even when you're not unsure about the solution :) – tom Oct 31 '11 at 0:59

Well you have by cross multiplying that,$$2(x+1)^{2} = 2x^{2} + x -3$$ $$\Longrightarrow 2x^{2}+4x+2 = 2x^{2} + x -3$$ $$\Longrightarrow 3x = -5 \Rightarrow x= -\frac{5}{3}$$
There’s nothing wrong with your basic approach, but you seem to have solved the equation $(2x+3)(x-1)=(x+1)(2x+2)$ incorrectly. After multiplying out both sides, you should have $$2x^2+x-3=2x^2+4x+2\;,$$ which simplifies progressively to \begin{align*} x-3&=4x+2\;,\\ -3&=3x+2\;,\\ -5&=3x\;,\text{ and}\\ x&=-\frac53. \end{align*}
Substituting this into the original equation yields \begin{align*} \frac{2\left(-\frac53\right)+3}{-\frac53+1} &\stackrel{?}= \frac{2\left(-\frac53\right)+2}{-\frac53-1}\\ \frac{-\frac13}{-\frac23} &\stackrel{?}= \frac{-\frac43}{-\frac83}\\ \frac12&=\frac12. \end{align*}